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CHAPTER 11 
 
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PROBLEM 11.1 
A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS 
coordinates are used to determine his displacement as a function of time: x= 0.5t3 + t2 + 2t where x and t are 
expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder 
when t = 5 seconds. 
 
SOLUTION 
Position: 3 20.5 2x t t t   
Velocity: 21.5 2 2
dx
v t t
dt
    
Acceleration: 3 2
dv
a t
dt
   
At 5 s,t     3 20.5 5 5 2 5x    97.5 ftx   
    21.5 5 2 5 2v    49.5 ft/sv   
  3 5 2a    217 ft/sa  
 
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PROBLEM 11.2 
The motion of a particle is defined by the relation 3 22 9 12 10,   x t t t where x and t are expressed in feet 
and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v  0. 
 
SOLUTION 
 3 22 9 12 10x t t t    
Differentiating, 2 26 18 12 6( 3 2)
6( 2)( 1)
      
  
dx
v t t t t
dt
t t
 
 12 18  
dv
a t
dt
 
So 0v at 1st and 2 s.t 
At 1s,t 1 2 9 12 10 15    x 1.000 st   
 1 12 18 6   a 1 15.00 ftx 
 21 6.00 ft/s a  
At 2 s,t 
 3 22 2(2) 9(2) 12(2) 10 14    x 2.00 st 
 2 14.00 ftx  
 2 (12)(2) 18 6a    
2
2 6.00 ft/sa 
 
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PROBLEM 11.3 
The vertical motion of mass A is defined by the relation 10 sin 2 15cos 2 100,x t t   
where x and t are expressed in mm and seconds, respectively. Determine (a) the position, 
velocity and acceleration of A when t  1 s, (b) the maximum velocity and acceleration of A. 
 
SOLUTION 
 10sin 2 15cos 2 100  x t t 
 20cos 2 30sin 2  
dx
v t t
dt
 
 40sin 2 60cos 2   
dv
a t t
dt
 
For trigonometric functions set calculator to radians: 
(a) At 1s.t 1 10sin 2 15cos 2 100 102.9x     1 102.9 mmx   
 1 20cos 2 30sin 2 35.6v     1 35.6 mm/s v  
 1 40sin 2 60cos 2 11.40    a 
2
1 11.40 mm/sa   
(b) Maximum velocity occurs when 0.a  
 40sin 2 60cos 2 0t t   
 
60
tan 2 1.5
40
t     
 12 tan ( 1.5) 0.9828t     and 0.9828   
 Reject the negative value. 2 2.1588t  
 1.0794 st  
 1.0794 st  for vmax
 
max 20cos(2.1588) 30sin(2.1588)
36.056
v  
  max 36.1 mm/sv   
 Note that we could have also used 
 2 2max 20 30 36.056v    
 by combining the sine and cosine terms. 
 For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms. 
 2 2 2max 40 60 72.1 mm/sa    
2
max 72.1 mm/sa   
 
so 
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PROBLEM 11.4 
A loaded railroad car is rolling at a constant velocity 
when it couples with a spring and dashpot bumper 
system. After the coupling, the motion of the car is 
defined by the relation 4.860 sin16tx e t where x and t 
are expressed in mm and seconds, respectively. 
Determine the position, the velocity and the 
acceleration of the railroad car when (a) t  0, 
(b) t  0.3 s. 
 
SOLUTION 
 4.860 sin16tx e t 
 
4.8 4.8
4.8 4.8
60( 4.8) sin16 60(16) cos16
288 sin16 960 cos16
t t
t t
dx
v e t e t
dt
v e t e t
 
 
   
  
 
 
4.8 4.8
4.8 4.8
4.8 4.8
1382.4 sin16 4608 cos16
4608 cos16 15360 sin16
13977.6 sin16 9216 cos16
t t
t t
t
dv
a e t e t
dt
e t e t
a e t e t
 
 
 
  
 
  
 
(a) At 0,t  0 0x  0 0 mmx   
 0 960 mm/sv  0 960 mm/sv   
 20 9216 mm/sa   
2
0 9220 mm/sa  
(b) At 0.3 s,t  4.8 1.44 0.23692
sin16 sin 4.8 0.99616
cos16 cos 4.8 0.08750
te e
t
t
  
  
 
 
 0.3 (60)(0.23692)( 0.99616) 14.16x     0.3 14.16 mmx   
 
0.3 (288)(0.23692)( 0.99616)
(960)(0.23692)(0.08750) 87.9
v   
  0.3 87.9 mm/sv   
 
0.3 (13977.6)(0.23692)( 0.99616)
(9216)(0.23692)(0.08750) 3108
a   
  20.3 3110 mm/sa   
 or 23.11 m/s  
 
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PROBLEM 11.5 
The motion of a particle is defined by the relation 4 3 26 2 12 3 3,x t t t t     where x and t are expressed in 
meters and seconds, respectively. Determine the time, the position, and the velocity when 0.a  
 
SOLUTION 
We have 4 3 26 2 12 3 3x t t t t     
Then 3 224 6 24 3
dx
v t t t
dt
     
and 272 12 24
dv
a t t
dt
    
When 0:a  2 272 12 24 12(6 2) 0t t t t      
or (3 2)(2 1) 0t t   
or 
2 1
s and s (Reject)
3 2
t t   0.667 st   
At 
2
s:
3
t  
4 3 2
2/3
2 2 2 2
6 2 12 3 3
3 3 3 3
x
                  
       
 2/3or 0.259 mx   
 
3 2
2/3
2 2 2
24 6 24 3
3 3 3
v
             
     
 2/3or 8.56 m/sv    
 
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PROBLEM 11.6 
The motion of a particle is defined by the relation 3 29 24 8,x t t t    where x and t are expressed in inches 
and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance 
traveled when the acceleration is zero. 
 
SOLUTION 
We have 3 29 24 8x t t t    
Then 23 18 24
dx
v t t
dt
    
and 6 18
dv
a t
dt
   
(a) When 0:v  2 23 18 24 3( 6 8) 0t t t t      
 ( 2)( 4) 0t t   
 2.00 s and 4.00 st t   
(b) When 0:a  6 18 0 or 3 st t   
 At 3 s:t  3 23 (3) 9(3) 24(3) 8x     3or 10.00 in.x   
First observe that 0 2 s:t  0v  
 2 s 3 s:t  0v  
Now 
At 0:t  0 8 in.x   
At 2 s:t  3 22 (2) 9(2) 24(2) 8 12 in.x      
 
Then 2 0
3 2
12 ( 8) 20 in.
| | |10 12| 2 in.
x x
x x
    
   
 
Total distance traveled (20 2)in.  Total distance 22.0 in. 
 
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PROBLEM 11.7 
A girl operates a radio-controlled model car in a vacant 
parking lot. The girl’s position is at the origin of the xy 
coordinate axes, and the surface of the parking lot lies in the 
x-y plane. She drives the car in a straight line so that the x 
coordinate is defined by the relation x(t) = 0.5t3 - 3t2 + 3t + 
2, where x and t are expressed in meters and seconds, 
respectively. Determine (a) when the velocity is zero, (b) 
the position and total distance travelled when the 
acceleration is zero. 
 
SOLUTION 
Position:   3 20.5 3 3 2x t t t t    
Velocity:   dxv t
dt
 
   21.5 6 3v t t t   
(a) Time when v=0 20 1.5 6 3t t   
 
  26 6 4 1.5 3
2 1.5
t
 


 0.586 s and 3.414 st t   
Acceleration:   dva t
dt
 
   3 6a t t  
Time when a=0 0 3 6t  
 2 st  
(b) Position at t=2 s      3 22 0.5 2 3 2 3 2 2x      
  2 0 mx   
To find total distance note that car changes direction at t=0.586 s 
 
Position at t=0 s      3 20 0.5 0 3 0 3 0 2x      
  0 2x  
Position at t=0.586 s      3 20.586 0.5 0.586 3 0.586 3 0.586 2x      
  0.586 2.828 mx  
Distances traveled: 
From t=0 to t=0.586 s:    0.586 0 0.828 mx x  
From t=0.586 to t=2 s:    2 0.586 2.828 mx x  
Total distance traveled 0.828 m 2.828 m  Total distance 3.656 m  
 
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PROBLEM 11.8 
The motion of a particle is defined by the relation  32 2 ,x t t   where x and t are expressed in feet and 
seconds, respectively. Determine (a) the two positions at which the velocity is zero, (b) the total distance 
traveled by the particle from t  0 to t  4 s.. 
 
 
SOLUTION 
Position:    32 2x t t t   
Velocity:   dxv t
dt
 
    22 3 2v t t t   
 
   2
2
2 3 4 4
 3 14 12
v t t t t
t t
   
   
 
Time when v(t)=0 20 3 14 12t t    
 
  
 
214 14 4 3 12
2 3
t
    


 
 1.131 s and 3.535 st t  
(a) Position at t=1.131 s      2 31.131 1.131 1.1.31 2x    
  1.131 1.935 ft.x   
 Position at t=3.535 s      2 33.535 3.535 3.535 2x    
  3.531 8.879 ft.x   
To find total distance traveled note that the particle changes direction at t=1.131 s and again at t=3.535 s. 
 Position at t=0 s      2 30 0 0 2x    
  0 8 ftx  
 Position at t=4 s      2 34 4 4 2x    
  4 8 ftx  
(b) Distances traveled: 
From t=0 to t=1.131 s:    1.131 0 6.065 ft.x x  
From t=1.131 to t=3.531 s:    3.535 1.131 6.944 ft.x x  
From t=3.531 to t=4 s:    4 3.535 0.879 ft.x x  
 
Total distance traveled 6.065 ft 6.944 ft 0.879 ft   Total distance 13.888 ft.  
 
 
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PROBLEM 11.9 
The brakes of a car are applied, causing it to slow down at a rate 
of 10 ft/s2. Knowing that the car stops in 100 ft, determine 
(a) how fast the car was traveling immediately before the brakes 
were applied, (b) the time required for the car to stop. 
 
SOLUTION 
 210 ft/s a 
(a) Velocity at 0.x 
 
0
0
0
2
0
2
0
10
( 10)
0 10 (10)(300)
2
6000
fx
v
f
dv
v a
dx
vdv dx
v
x
v
  
  
    

 
 0 77.5 ft/sv  
(b) Time to stop. 
 
0
0
0
0
0
10
10
0 10
77.5
10 10
ft
v
f
f
dv
a
dx
dv dt
v t
v
t
  
  
  
 
 
 7.75 sft   
 
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PROBLEM 11.10 
The acceleration of a particle is defined by the relation 0.23 ,ta e where a and t are expressed in 2ft/s and 
seconds, respectively. Knowing that x 0 and v 0 at t 0, determine the velocity and position of the 
particle when t 0.5 s. 
 
SOLUTION 
Acceleration: 0.2 23 ft/sta e 
Given : 0 00 ft/s, 0 ftv x  
Velocity: 
dv
a dv adt
dt
   
 
 
0 0
0.2 0.2
0
0
0.2
3 0= 15
15 1 ft/s
v t
v
t t
t t
o
t
dv adt
v v e dt v e
v e
 


    
 
 
 
Position: 
dx
v dx vdt
dt
   
    
 
0 0
0.2 0.2
0
0
0.2
15 1 x 0=15 5
15 5 75 ft
x t
x
t t
t t
o
t
dx vdt
x x e dt t e
x t e
 


     
  
 
 
Velocity at t=0.5 s  0.2 0.515 1 ft/sv e   
  0.5 1.427 ft/sv   
Position at t=0.5 s  0.2 0.515 0.5 5 75 ftx e    
  0.5 0.363 ft.x   
 
 
 
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PROBLEM 11.11 
The acceleration of a particle is directly proportional to the square of the time t. When 0,t  the particle is 
at 24 m.x  Knowing that at 6 s, 96 mt x  and 18 m/s,v  express x and v in terms of t. 
 
SOLUTION 
We have 2 constanta kt k  
Now 2
dv
a kt
dt
  
At 6 s, 18 m/s:t v  2
18 6
v t
dv kt dt  
or 3
1
18 ( 216)
3
v k t   
or 3
1
18 ( 216)(m/s)
3
v k t   
Also 3
1
18 ( 216)
3
dx
v k t
dt
    
At 0, 24 m:t x  3
24 0
1
18 ( 216)
3
x t
dx k t dt
       
or 4
1 1
24 18 216
3 4
x t k t t
     
 
 
Now 
At 6 s, 96 m:t x  4
1 1
96 24 18(6) (6) 216(6)
3 4
k
      
 
or 4
1
m/s
9
k  
Then 4
1 1 1
24 18 216
3 9 4
x t t t
       
  
 
or 4
1
( ) 10 24
108
x t t t    
and 3
1 1
18 ( 216)
3 9
v t
    
 
 
or 3
1
( ) 10
27
v t t   
 
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PROBLEM 11.12 
The acceleration of a particle is defined by the relation 2.a kt (a) Knowing that 8 m/s v when t  0 
and that 8 m/s v when 2 s,t determine the constant k. (b) Write the equations of motion, knowing also 
that 0x  when 2 s.t 
 
SOLUTION 
 2
2
a kt
dv
a kt
dt

 
 (1) 
0, 8 m/s  t v and 2 s, 8 ft/s  t v 
(a) 
8 2
2
8 0
dv kt dt

  
 3
1
8 ( 8) (2)
3
   k 46.00 m/sk  
(b) Substituting 46 m/s into (1)k  
 26 
dv
a t
dt
 26a t  
0, 8 m/s:  t v 2
8 0
6
v tdv t dt

  
 3
1
( 8) 6( )
3
  v t 32 8 v t  
 32 8  
dx
v t
dt
 
2 s, 0: t x 3 4
0 2
2
1
(2 8) ; 8
2
t
x t
dx t dt x t t     
 
4 4
4
1 1
8 (2) 8(2)
2 2
1
8 8 16
2
            
   
x t t
x t t 4
1
8 8
2
  x t t 
 
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PROBLEM 11.13 
A Scotch yoke is a mechanism that transforms the circular motion of a 
crank into the reciprocating motion of a shaft (or vice versa). It has been 
used in a number of different internal combustion engines and in control 
valves. In the Scotch yoke shown, the acceleration of Point A is defined by 
the relation a 1.8sin kt, where a and t are expressed in m/s2 and seconds, 
respectively, and k 3 rad/s. Knowing that x 0 and v 0.6 m/s when t 0, 
determine the velocity and position of Point A when t 0.5 s. 
 
SOLUTION 
Acceleration: 21.8sin m/sa kt  
Given : 0 00.6 m/s, 0, 3 rad/sv x k   
Velocity: 
0 0
 
v t
v
dv
a dv adt dv adt
dt
      
 0 0 0 0
1.8
1.8 sin cos
tt t
v v a dt kt dt kt
k
      
 
1.8
0.6 (cos 1) 0.6cos 0.6
3
    v kt kt 
 0.6cos m/sv kt 
Position: 
0 0
 
x t
x
dx
v dx vdt dx vdt
dt
      
 0 0 0 0
0.6
0.6 cos sin
tt t
x x v dt kt dt kt
k
     
 
0.6
0 (sin 0) 0.2sin
3
   x kt kt 
 0.2sin mx kt 
When t =0.5 s, (3)(0.5) 1.5 rad kt 
 0.6cos1.5 0.0424 m/sv   42.4 mm/s v   
 0.2sin1.5 0.1995 mx   199.5 mm x   
 
 
 
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PROBLEM 11.14 
For the scotch yoke mechanism shown, the acceleration of Point A is defined by the 
relation 1.08sin 1.44cos ,a kt kt   where a and t are expressed in m/s2 and 
seconds, respectively, and k = 3 rad/s. Knowing that x = 0.16 m and v = 0.36 m/s 
when t = 0, determine the velocity and position of Point A when t = 0.5 s. 
 
SOLUTION 
Acceleration: 21.08sin 1.44cos m/sa kt kt   
Given : 0 00.36 m/s, 0.16, 3 rad/sv x k   
Velocity: 
0 0
 
v t
v
dv
a dv adt dv adt
dt
      
Integrate: 0 0 0
0 0
1.08 sin 1.44 cos
1.08 1.44
0.36 cos sin
1.08 1.44
(cos3 1) (sin 3 0)
3 3
0.36cos3 0.36 0.48sin 3
0.36cos3 0.48sin 3 m/s
t t
t t
v v kt dt kt dt
v kt kt
k k
t t
t t
v t t
   
  
   
  
 
  
Evaluate at 0.5 st  0.36cos1.5 0.36 0.48sin1.5   453 mm/s v    
Position: 
0 0
 
x t
x
dx
v dx vdt dx vdt
dt
      
 
0 0 0 0
0 0
0.36 cos 0.48 sin
0.36 0.48
0.16 sin cos
0.36 0.48
(sin 3 0) (cos3 1)
3 3
0.12sin 3 0.16cos3 0.16
0.12sin 3 0.16cos3 m
t t t
t t
x x v dt kt dt kt dt
x kt kt
k k
t t
t t
x t t
   
  
   
  
 
  
 
Evaluate at 0.5 st  0.12sin1.5 0.16cos1.5 0.1310 mx    131.0 mm x   
 
 
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PROBLEM 11.15 
A piece of electronic equipment that is surrounded by 
packing material is dropped so that it hits the ground with a 
speed of 4 m/s. After contact the equipment experiences an 
acceleration of ,a kx  where k is a constant and x is the 
compression of the packing material. If the packing material 
experiences a maximum compression of 20 mm, determine 
the maximum acceleration of the equipment. 
 
SOLUTION 
 
vdv
a kx
dx
   
Separate and integrate. 
 
0 0
2 2 2 2
0
0
1 1 1 1
2 2 2 2
f f
f
v x
v
x
f f
vdv kx dx
v v kx kx
 
    
 
 
Use 0 4 m/s, 0.02 m, and 0.f fv x v   Solve for k. 
 2 2 2
1 1
0 (4) (0.02) 40,000 s
2 2
   k k 
Maximum acceleration. 
 2max max : ( 40,000)(0.02) 800 m/sa kx     
 2800 m/sa  
 
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PROBLEM 11.16 
A projectile enters a resisting medium at 0x  with an initial velocity 
v0 900 ft/s and travels 4 in. before coming to rest. Assuming that the 
velocity of the projectile is defined by the relation 0 ,v v kx  where v is 
expressed in ft/s and x is in feet, determine (a) the initial acceleration of the 
projectile, (b) the time required for the projectile to penetrate 3.9 in. into the 
resisting medium. 
 
SOLUTION 
First note 
When 
4
ft, 0:
12
x v  
4
0 (900 ft/s) ft
12
k
    
 
 
or 
1
2700
s
k  
(a) We have 0v v kx  
 Then 0( )
dv d
a v kx kv
dt dt
     
 or 0( )a k v kx   
 At 0:t  
1
2700 (900 ft/s 0)
s
a   
 or 6 20 2.43 10 ft/sa     
(b) We have 0
dx
v v kx
dt
   
 At 0, 0:t x  
0 0
0
x tdx
dt
v kx

  
 or 0 0
1
[ln( )]xv kx t
k
   
 or 
0
0
0
1 1 1
ln ln
1 kv
v
t
k v kx k x
  
         
 
 When 3.9 in.:x   2700 1/s 3.91 900 ft/s 12s
11
ln
1 ft2700
t
 
   
 
 or 31.366 10 st    
 
 
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PROBLEM 11.17 
The acceleration of a particle is defined by the relation / .a k x  It has been experimentally determined that 
15 ft/sv  when 0.6 ftx  and that 9 ft/sv  when 1.2 ft.x  Determine (a) the velocity of the particle 
when 1.5 ft,x  (b) the position of the particle at which its velocity is zero. 
 
SOLUTION 
 
vdv k
a
dx x

  
Separate and integrate using 0.6 ft, 15 ft/s.x v  
 
15 0.6
2
15 0.6
2 2
1
ln
2
1 1
(15) ln
2 2 0.6
v x
xv
dx
vdv k
x
v k x
x
v k
 
 
     
 
 
 (1) 
When 9 ft/s, 1.2 ftv x  
 2 2
1 1 1.2
(9) (15) ln
2 2 0.6
k
     
 
 
Solve for k. 
 2 2103.874 ft /sk  
(a) Velocity when 65 ft.x 
 Substitute 2 2103.874 ft /s and 1.5 ftk x  into (1). 
 2 2
1 1 1.5
(15) 103.874 ln
2 2 0.6
v
     
 
 
 5.89 ft/sv   
(b) Position when for 0,v  
 
210 (15) 103.874 ln
2 0.6
ln 1.083
0.6
x
x
     
 
   
 
 
   1.772 ftx   
 
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PROBLEM 11.18 
A brass (nonmagnetic) block A and a steel magnet B are in equilibrium in a brass 
tube under the magnetic repelling force of another steel magnet C located at a 
distance x  0.004m from B. The force is inversely proportional to the square of 
the distance between B and C. If block A is suddenly removed, the acceleration 
of block B is 29.81 / ,a k x   where a and x are expressed in m/s2 and m, 
respectively, and 4 3 24 10 m /s .k   Determine the maximum velocity and 
acceleration of B. 
 
SOLUTION 
The maximum velocity occurs when 0.a 
2
0 9.81
m
k
x
   
 
4
2 6 24 10 40.775 10 m 0.0063855 m
9.81 9.81m m
k
x x

     
The acceleration is given as a function of x. 
 
2
9.81   
dv k
v a
dx x
 
Separate variables and integrate: 
 
0 0
2
20
2
0
0
9.81
9.81
1 1 1
9.81( )
2
v x x
x x
k dx
vdv dx
x
dx
vdv dx k
x
v x x k
x x
  
  
 
     
 
   
 
2
0
0
4
2 2
1 1 1
9.81( )
2
1 1
9.81(0.0063855 0.004) (4 10 )
0.0063855 0.004
0.023402 0.037358 0.013956 m /s
m m
m
v x x k
x x

 
     
 
       
 
   
 
Maximum velocity: 0.1671 m/smv 167.1 mm/smv  
The maximum acceleration occurs when x is smallest, that is, 0.004 m.x 
 
4
2
4 10
9.81
(0.004)
ma

   215.19 m/sma  
 
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PROBLEM 11.19 
Based on experimental observations, the acceleration of a particle is defined by the relation (0.1a    sin x/b), 
where a and x are expressed in m/s2 and meters, respectively. Knowing that 0.8 mb  and that 1 m/sv  
when 0,x  determine (a) the velocity of the particle when 1 m,x   (b) the position where the velocity is 
maximum, (c) the maximum velocity. 
 
SOLUTION 
We have 0.1 sin
0.8
dv x
v a
dx
     
 
 
When 0, 1 m/s:x v  
1 0
0.1 sin
0.8
v x x
vdv dx
    
   
or 2
0
1
( 1) 0.1 0.8 cos
2 0.8
x
x
v x
      
 
or 2
1
0.1 0.8 cos 0.3
2 0.8
x
v x    
(a) When 1 m:x   2
1 1
0.1( 1) 0.8 cos 0.3
2 0.8
v

     
 or 0.323 m/sv    
(b) When max, 0:v v a  0.1 sin 00.8
x    
 
 
 or 0.080134 mx   0.0801 mx    
(c) When 0.080134 m:x   
 
2
max
2 2
1 0.080134
0.1( 0.080134) 0.8 cos 0.3
2 0.8
0.504 m /s
v

    

 
 or max 1.004 m/sv   
 
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PROBLEM 11.20 
A spring AB is attached to a support at A and to a collar. The 
unstretched length of the spring is l. Knowing that the collar is 
released from rest at 0x x and has an acceleration defined by 
the relation 2 2100( / )   a x lx l x , determine the velocity of 
the collar as it passes through Point C. 
 
SOLUTION 
Since a is function of x, 
2 2
100
 
      
dv lx
a v x
dx l x
 
Separate variables and integrate: 
 
0 0
0
2 2
100
fv
v x
lx
vdv x dx
l x
 
     
  
 
0
0
2
2 2 2 2
0
1 1
100
2 2 2
 
      
 
f
x
x
v v l l x 
 
2
2 2 2 20
0
1
0 100
2 2
 
        
 
f
x
v l l l x 
 
2 2 2 2 2 2
0 0
2 2 2
0
1 100
( 2 )
2 2
100
( )
2
fv l x l l l x
l x l
     
  
 
 2 2010( )fv l x l    
 
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PROBLEM 11.21 
The acceleration of a particle is defined by the relation  1 ,xa k e  where k is a constant. Knowing that 
the velocity of the particle is 9v   m/s when 3x   m and that the particle comes to rest at the origin, 
determine (a) the value of k, (b) the velocity of the particle when 2x   m. 
 
SOLUTION 
Acceleration:  1 xa k e  
Given : at 3x   m, 9v  m/s 
 at 0x  m, 0v  m/s 
   k 1 x
adx vdv
e dx vdv

 
 
Integrate using x=-3 m and v=9 m/s as the lower limits of the integrals 
 
 
 
3 9
2
93
 k 1
1
2
x v
x
x v
x
e dx vdv
k x e v




 
 
 
 
Velocity:   3 2 21 13 (9)
2 2
xk x e e v      (1) 
(a) Now substitute v=0 m/s and x=0 m into (1) and solve for k 
   0 3 2 21 10 3 0 (9)
2 2
k e e      
 2 22.52 m /sk   
(b) Find velocity when x= -2 m using the equation (1) and the value of k 
   2 3 2 21 12.518 2 3 (9)
2 2
e e v       
 4.70 m/sv   
 
 
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PROBLEM 11.22 
Starting from x  0 with no initial velocity, a particle is given an acceleration 20.1 16,a v  where a 
and v are expressed in ft/s2 and ft/s, respectively. Determine (a) the position of the particle when v  
3ft/s, (b) the speed and acceleration of the particle when x  4 ft. 
 
SOLUTION 
 2 1/20.1( 16)
vdv
a v
dx
   (1) 
Separate and integrate. 
 
20 0
0.1
16
v xvdv
dx
v


  
 
2 1/2
0
2 1/2
2 1/2
( 16) 0.1
( 16) 4 0.1
10[( 16) 4]
v
v x
v x
x v
 
  
   (2) 
(a) 3 ft/s.v  
 2 1/210[(3 16) 4]x    10.00 ftx   
(b) 4 ft.x  
 From (2), 2 1/2( 16) 4 0.1 4 (0.1)(4) 4.4v x      
 
2
2 2 2
16 19.36
3.36ft /s
v
v
 
 1.833 ft/sv   
 From (1), 2 1/20.1(1.833 16)a   20.440 ft/sa   
 
 
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PROBLEM 11.23 
A ball is dropped from a boat so that it strikes the surface of a 
lake with a speed of 16.5 ft/s. While in the water the ball 
experiences an acceleration of 10 0.8 ,a v  where a and v 
are expressed in ft/s2 and ft/s, respectively. Knowing the 
ball takes 3 s to reach the bottom of the lake, determine (a) the 
depth of the lake, (b) the speed of the ball when it hits the 
bottom of the lake. 
 
SOLUTION 
 10 0.8
dv
a v
dt
   
Separate and integrate: 
 
0 010 0.8
v t
v
dv
dt
v

  
 0
0
1
ln(10 0.8 )
0.8
10 0.8
ln 0.8
10 0.8
v
v
v t
v
t
v
  
 
   
 
 0.8010 0.8 (10 0.8 )
tv v e   
or 0.80
0.8
0
0.8 10 (10 0.8 )
12.5 (12.5 )
t
t
v v e
v v e


  
  
 
With 0 16.5ft/sv  
0.812.5 4 tv e  
Integrate to determine x as a function of t. 
 0.812.5 4 t
dx
v e
dt
   
 0.8
0 0
(12.5 4 )
x t
tdx e dt   
 
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https://testbanku.eu/Solution-Manual-for-Vector-Mechanics-for-Engineers-Dynamics-11th-Edition-by-BeerPROBLEM 11.23 (Continued) 
 0.8 0.8
0
12.5 5 12.5 5 5
tt tx t e t e      
(a) At 35 s,t  
 2.412.5(3) 5 5 42.046 ftx e    42.0 ftx   
(b) 2.412.5 4 12.863 ft/sv e   12.86 ft/sv   
 
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PROBLEM 11.24 
The acceleration of a particle is defined by the relation ,a k v  where k is a constant. Knowing that x  0 
and v  81 m/s at t  0 and that v  36 m/s when x  18 m, determine (a) the velocity of the particle when 
x  20 m, (b) the time required for the particle to come to rest. 
 
SOLUTION 
(a) We have 
dv
v a k v
dx
   
 so that v dv kdx  
 When 0, 81 m/s:x v  
81 0
v x
v dv kdx   
 or 3/2 81
2
[ ]
3
vv kx  
 or 3/2
2
[ 729]
3
v kx   
 When 18 m, 36 m/s:x v  3/2
2
(36 729) (18)
3
k   
 or 219 m/sk  
 Finally 
 When 20 m:x  3/2
2
( 729) 19(20)
3
v    
 or 3/2 159v  29.3 m/sv   
(b) We have 19
dv
a v
dt
   
 At 0, 81 m/s:t v  
81 0
19
v tdv
dt
v
   
 or 812[ ] 19
vv t  
 or 2( 9) 19v t   
 When 0:v  2( 9) 19t   
 or 0.947 st  
 
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PROBLEM 11.25 
The acceleration of a particle is defined by the relation 2.5,a kv  where k is a constant. The particle starts at 
0x  with a velocity of 16 mm/s, and when 6x  mm the velocity is observed to be 4 mm/s. Determine 
(a) the velocity of the particle when 5x  mm, (b) the time at which the velocity of the particle is 9 mm/s. 
 
SOLUTION 
Acceleration: 2.5a kv  
Given : at t=0, 0x  mm, 16v  mm/s 
 at 6x  mm, 4v  mm/s 
 
2.5
adx vdv
kv dx vdv

 
 
Separate variables 3 2kdx v dv  
 
Integrate using x=-0 m and v=16 mm/s as the lower limits of the integrals 
 
3 2
0 16
1 2
16
0
 -k
2
x v
x
v
dx v dv
kx v



  
 
 
Velocity and position: 1 2
1
2
2
kx v  (1) 
Now substitute v=4 mm/s and x=6 mm into (1) and solve for k 
 
  1 2
3 2 1 2
1
6 4
2
0.0833 mm
k
k s


 

 
 
(a) Find velocity when x= 5 mm using the equation (1) and the value of k 
   1 2 15 2
2
k v  
 4.76 mm/sv   
(b) or 
dv
a adt dv
dt
  
 or 
dv
a adt dv
dt
  
Separate variables 2.5kdt v dv  
 
 
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PROBLEM 11.25 (Continued) 
 
Integrate using t=0 and v=16 mm/s as the lower limits of the integrals 
 
5 2
0 16
3 2
160
 -k
2
3
t v
t v
dt v dv
kt v



  
 
 
Velocity and time: 3 2
2 1
3 96
kt v  (2) 
Find time when v= 9 mm/s using the equation (2) and the value of k 
   3 22 19
3 96
kt
  
 0.171 st   
 
 
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Problem 11.26 
A human powered vehicle (HPV) team wants to model the 
acceleration during the 260 m sprint race (the first 60 m is called a 
flying start) using a = A – Cv2, where a is acceleration in m/s2 and v 
is the velocity in m/s. From wind tunnel testing, they found that 
C = 0.0012 m-1. Knowing that the cyclist is going 100 km/h at the 
260 meter mark, what is the value of A? 
 
SOLUTION 
Acceleration: 2 2 m/sa A Cv  
Given : 10.0012 mC  , 100 km/hr when 260 mf fv x  
 Note: 100 km/hr = 27.78 m/s 
  2 
adx vdv
A Cv dx vdv

 
 
Separate variables 
2 
vdv
dx
A Cv


 
Integrate starting from rest and traveling a distance xf with a final velocity vf. 
 
 
   
2
0 0
2
0
0
2
2
 
 
1
ln
2
1 1
ln ln
2 2
1
ln
2
f f
f
f
x v
v
x
f f
f
f
vdv
dx
A Cv
x A Cv
C
x A v A
C C
A
x
C A Cv


  
   
 
 
  
 
 
Next solve for A 
22
2
1
f
f
Cx
f
Cx
Cv e
A
e
 

 
Now substitute values of C, xf and vf and solve for A 
21.995 m/sA   
 
 
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